Recently, I found the following three (conjectural) identities for $\pi^2$: $$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3,\tag{1}$$ $$\sum_{k=1}^\infty\frac{(42k^2-23k+3)16^k}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^2}2,\tag{2}$$ $$\sum_{k=1}^\infty\frac{(92k^2-61k+9)64^k}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=8\pi^2.\tag{3}$$ The converging rates of the series in $(1)$-$(3)$ are $4/729$, $1/64$ and $4/27$, respectively. One can easily check the identities $(1)$-$(3)$ numerically.

Let $P(k)=145k^2-104k+18$. Motivated by $(1)$, I also conjecture the following identities involving harmonic numbers:

$$\sum_{k=1}^\infty\frac{6P(k)(H_{3k-1}-H_{k-1})-232k+89}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=18\zeta(3),\tag{4}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{2k-1}-H_{k-1})-\frac{3(58k^2-40k+7)}{2(2k-1)}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\zeta(3),\tag{5}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{3k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{17k+32}{9k}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{7\pi^4}{180},\tag{6}$$ $$\sum_{k=1}^\infty\frac{P(k)(297H_{3k-1}^{(2)}-192H_{2k-1}^{(2)}-978H_{k-1}^{(2)})+\frac{27(180k^2+12k-35)}{(2k-1)^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{167}{20}\pi^4,\tag{7}$$ and $$\sum_{k=1}^\infty\frac{P(k)H_{k-1}^{(3)}+\frac{28(2k-1)}{17k^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{528\zeta(5)-46\pi^2\zeta(3)}{17},\tag{8}$$ where $H_n:=\sum_{0<j\le n}\frac1j$ and $H_n^{(m)}:=\sum_{0<j\le n}\frac1{j^m}$ for $m=2,3,\ldots$.

Let $Q(k)=(3k-1)(14k-3)=42k^2-23k+3$. Motivated by $(2)$, I conjecture the following identities: $$\sum_{k=1}^\infty\frac{16^k\left(6Q(k)(H_{2k-1}-H_{k-1})-\frac{196k^2-100k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=2\pi^2\log2-7\zeta(3),\tag{9}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(H_{4k-1}-H_{2k-1})-\frac{28k^2-76k+19}{12(2k-1)}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{2\pi^2\log2+35\zeta(3)}{12},\tag{10}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-H_{2k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{6k+1}{k}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{5\pi^4}{24},\tag{11}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-5H_{2k-1}^{(2)}-3H_{k-1}^{(2)})+\frac{32k(3k-1)}{(2k-1)^2}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^4}{6}.\tag{12}$$

Let $R(k)=92k^2-61k+9$. Inspired by $(3)$, I also conjecture the following identities: $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{3k-1}-5H_{k-1})-60k+21\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=48\pi^2\log2-56\zeta(3),\tag{13}$$ $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{2k-1}-2H_{k-1})-\frac{124k^2-79k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=112\zeta(3),\tag{14}$$ and $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(6H_{4k-1}-H_{k-1})-\frac{344k^2-281k+59}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=560\zeta(3).\tag{15}$$

QUESTION. Can one prove the new identities $(1)-(15)$ via known methods (including the WZ method and hypergeometric series identies)?

Your comments are welcome!

Recently, I found the following three (conjectural) identities for $\pi^2$: $$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3,\tag{1}$$ $$\sum_{k=1}^\infty\frac{(42k^2-23k+3)16^k}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^2}2,\tag{2}$$ $$\sum_{k=1}^\infty\frac{(92k^2-61k+9)64^k}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=8\pi^2.\tag{3}$$ The converging rates of the series in $(1)$-$(3)$ are $4/729$, $1/64$ and $4/27$, respectively. One can easily check the identities $(1)$-$(3)$ numerically.

Let $P(k)=145k^2-104k+18$. Motivated by $(1)$, I also conjecture the following identities involving harmonic numbers:

$$\sum_{k=1}^\infty\frac{6P(k)(H_{3k-1}-H_{k-1})-232k+89}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=18\zeta(3),\tag{4}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{2k-1}-H_{k-1})-\frac{3(58k^2-40k+7)}{2(2k-1)}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\zeta(3),\tag{5}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{3k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{17k+32}{9k}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{7\pi^4}{180},\tag{6}$$ $$\sum_{k=1}^\infty\frac{P(k)(297H_{3k-1}^{(2)}-192H_{2k-1}^{(2)}-978H_{k-1}^{(2)})+\frac{27(180k^2+12k-35)}{(2k-1)^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{167}{20}\pi^4,\tag{7}$$ and $$\sum_{k=1}^\infty\frac{P(k)H_{k-1}^{(3)}+\frac{28(2k-1)}{17k^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{528\zeta(5)-46\pi^2\zeta(3)}{17},\tag{8}$$ where $H_n:=\sum_{0<j\le n}\frac1j$ and $H_n^{(m)}:=\sum_{0<j\le n}\frac1{j^m}$ for $m=2,3,\ldots$.

Let $Q(k)=(3k-1)(14k-3)=42k^2-23k+3$. Motivated by $(2)$, I conjecture the following identities: $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(H_{2k-1}-H_{k-1})-\frac{196k^2-100k+13}{6(2k-1)}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{2\pi^2\log2-7\zeta(3)}6,\tag{9}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(H_{4k-1}-H_{2k-1})-\frac{28k^2-76k+19}{12(2k-1)}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{2\pi^2\log2+35\zeta(3)}{12},\tag{10}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-H_{2k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{6k+1}{k}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{5\pi^4}{24},\tag{11}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-5H_{2k-1}^{(2)}-3H_{k-1}^{(2)})+\frac{32k(3k-1)}{(2k-1)^2}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^4}{6}.\tag{12}$$

Let $R(k)=92k^2-61k+9$. Inspired by $(3)$, I also conjecture the following identities: $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{3k-1}-5H_{k-1})-60k+21\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=48\pi^2\log2-56\zeta(3),\tag{13}$$ $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{2k-1}-2H_{k-1})-\frac{124k^2-79k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=112\zeta(3),\tag{14}$$ and $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(6H_{4k-1}-H_{k-1})-\frac{344k^2-281k+59}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=560\zeta(3).\tag{15}$$

QUESTION. Can one prove the new identities $(1)-(15)$ via known methods (including the WZ method and hypergeometric series identies)?

Your comments are welcome!

Recently, I found the following three (conjectural) identities for $\pi^2$: $$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3,\tag{1}$$ $$\sum_{k=1}^\infty\frac{(42k^2-23k+3)16^k}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^2}2,\tag{2}$$ $$\sum_{k=1}^\infty\frac{(92k^2-61k+9)64^k}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=8\pi^2.\tag{3}$$ The converging rates of the series in $(1)$-$(3)$ are $4/729$, $1/64$ and $4/27$, respectively. One can easily check the identities $(1)$-$(3)$ numerically.

Let $P(k)=145k^2-104k+18$. Motivated by $(1)$, I also conjecture the following identities involving harmonic numbers:

$$\sum_{k=1}^\infty\frac{6P(k)(H_{3k-1}-H_{k-1})-232k+89}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=18\zeta(3),\tag{4}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{2k-1}-H_{k-1})-\frac{3(58k^2-40k+7)}{2(2k-1)}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\zeta(3),\tag{5}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{3k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{17k+32}{9k}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{7\pi^4}{180},\tag{6}$$ $$\sum_{k=1}^\infty\frac{P(k)(297H_{3k-1}^{(2)}-192H_{2k-1}^{(2)}-978H_{k-1}^{(2)})+\frac{27(180k^2+12k-35)}{(2k-1)^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{167}{20}\pi^4,\tag{7}$$ and $$\sum_{k=1}^\infty\frac{P(k)H_{k-1}^{(3)}+\frac{28(2k-1)}{17k^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{528\zeta(5)-46\pi^2\zeta(3)}{17},\tag{8}$$ where $H_n:=\sum_{0<j\le n}\frac1j$ and $H_n^{(m)}:=\sum_{0<j\le n}\frac1{j^m}$ for $m=2,3,\ldots$.

Let $Q(k)=(3k-1)(14k-3)=42k^2-23k+3$. Motivated by $(2)$, I conjecture the following identities: $$\sum_{k=1}^\infty\frac{16^k\left(6Q(k)(H_{2k-1}-H_{k-1})-\frac{196k^2-100k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}62}=2\pi^2\log2-7\zeta(3),\tag{9}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(H_{4k-1}-H_{2k-1})-\frac{28k^2-76k+19}{12(2k-1)}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}62}=\frac{2\pi^2\log2+35\zeta(3)}{12},\tag{10}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-H_{2k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{6k+1}{k}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}62}=\frac{5\pi^4}{24},\tag{11}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-5H_{2k-1}^{(2)}-3H_{k-1}^{(2)})+\frac{32k(3k-1)}{(2k-1)^2}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}62}=\frac{\pi^4}{6}.\tag{12}$$

Let $R(k)=92k^2-61k+9$. Inspired by $(3)$, I also conjecture the following identities: $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{3k-1}-5H_{k-1})-60k+21\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=48\pi^2\log2-56\zeta(3),\tag{13}$$ $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{2k-1}-2H_{k-1})-\frac{124k^2-79k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=112\zeta(3),\tag{14}$$ and $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(6H_{4k-1}-H_{k-1})-\frac{344k^2-281k+59}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=560\zeta(3).\tag{15}$$

QUESTION. Can one prove the new identities $(1)-(15)$ via known methods (including the WZ method and hypergeometric series identies)?

Your comments are welcome!

Recently, I found the following three (conjectural) identities for $\pi^2$: $$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3,\tag{1}$$ $$\sum_{k=1}^\infty\frac{(42k^2-23k+3)16^k}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^2}2,\tag{2}$$ $$\sum_{k=1}^\infty\frac{(92k^2-61k+9)64^k}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=8\pi^2.\tag{3}$$ The converging rates of the series in $(1)$-$(3)$ are $4/729$, $1/64$ and $4/27$, respectively. One can easily check the identities $(1)$-$(3)$ numerically.

Let $P(k)=145k^2-104k+18$. Motivated by $(1)$, I also conjecture the following identities involving harmonic numbers:

$$\sum_{k=1}^\infty\frac{6P(k)(H_{3k-1}-H_{k-1})-232k+89}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=18\zeta(3),\tag{4}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{2k-1}-H_{k-1})-\frac{3(58k^2-40k+7)}{2(2k-1)}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\zeta(3),\tag{5}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{3k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{17k+32}{9k}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{7\pi^4}{180},\tag{6}$$ $$\sum_{k=1}^\infty\frac{P(k)(297H_{3k-1}^{(2)}-192H_{2k-1}^{(2)}-978H_{k-1}^{(2)})+\frac{27(180k^2+12k-35)}{(2k-1)^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{167}{20}\pi^4,\tag{7}$$ and $$\sum_{k=1}^\infty\frac{P(k)H_{k-1}^{(3)}+\frac{28(2k-1)}{17k^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{528\zeta(5)-46\pi^2\zeta(3)}{17},\tag{8}$$ where $H_n:=\sum_{0<j\le n}\frac1j$ and $H_n^{(m)}:=\sum_{0<j\le n}\frac1{j^m}$ for $m=2,3,\ldots$.

Let $Q(k)=(3k-1)(14k-3)=42k^2-23k+3$. Motivated by $(2)$, I conjecture the following identities: $$\sum_{k=1}^\infty\frac{16^k\left(6Q(k)(H_{2k-1}-H_{k-1})-\frac{196k^2-100k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=2\pi^2\log2-7\zeta(3),\tag{9}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(H_{4k-1}-H_{2k-1})-\frac{28k^2-76k+19}{12(2k-1)}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{2\pi^2\log2+35\zeta(3)}{12},\tag{10}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-H_{2k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{6k+1}{k}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{5\pi^4}{24},\tag{11}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-5H_{2k-1}^{(2)}-3H_{k-1}^{(2)})+\frac{32k(3k-1)}{(2k-1)^2}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^4}{6}.\tag{12}$$

Let $R(k)=92k^2-61k+9$. Inspired by $(3)$, I also conjecture the following identities: $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{3k-1}-5H_{k-1})-60k+21\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=48\pi^2\log2-56\zeta(3),\tag{13}$$ $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{2k-1}-2H_{k-1})-\frac{124k^2-79k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=112\zeta(3),\tag{14}$$ and $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(6H_{4k-1}-H_{k-1})-\frac{344k^2-281k+59}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=560\zeta(3).\tag{15}$$

QUESTION. Can one prove the new identities $(1)-(15)$ via known methods (including the WZ method and hypergeometric series identies)?

Your comments are welcome!

Recently, I found the following two (conjectural) identities for $\pi^2$: $$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3\tag{1}$$ and $$\sum_{k=1}^\infty\frac{(92k^2-61k+9)64^k}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=8\pi^2.\tag{2}$$ The converging rates of $(1)$ and $(2)$ are $4/729$ and $4/27$, respectively. One can easily check $(1)$ and $(2)$ numerically.

Let $P(k)=145k^2-104k+18$. Motivated by $(1)$, I also conjecture the following identities involving harmonic numbers:

$$\sum_{k=1}^\infty\frac{6P(k)(H_{3k-1}-H_{k-1})-232k+89}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=18\zeta(3),\tag{3}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{2k-1}-H_{k-1})-\frac{3(58k^2-40k+7)}{2(2k-1)}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\zeta(3),\tag{4}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{3k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{17k+32}{9k}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{7\pi^4}{180},\tag{5}$$ $$\sum_{k=1}^\infty\frac{P(k)(297H_{3k-1}^{(2)}-192H_{2k-1}^{(2)}-978H_{k-1}^{(2)})+\frac{27(180k^2+12k-35)}{(2k-1)^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{167}{20}\pi^4,\tag{6}$$ and $$\sum_{k=1}^\infty\frac{P(k)H_{k-1}^{(3)}+\frac{28(2k-1)}{17k^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{528\zeta(5)-46\pi^2\zeta(3)}{17},\tag{7}$$ where $H_n:=\sum_{0<j\le n}\frac1j$ and $H_n^{(m)}:=\sum_{0<j\le n}\frac1{j^m}$ for $m=2,3,\ldots$.

Let $Q(k)=92k^2-61k+9$. Inspired by $(2)$, I also conjecture the following identities: $$\sum_{k=1}^\infty\frac{64^k\left(Q(k)(3H_{3k-1}-5H_{k-1})-60k+21\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=48\pi^2\log2-56\zeta(3),\tag{8}$$ $$\sum_{k=1}^\infty\frac{64^k\left(Q(k)(3H_{2k-1}-2H_{k-1})-\frac{124k^2-79k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=112\zeta(3),\tag{9}$$ and $$\sum_{k=1}^\infty\frac{64^k\left(Q(k)(6H_{4k-1}-H_{k-1})-\frac{344k^2-281k+59}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=560\zeta(3).\tag{10}$$

QUESTION. Can one prove the new identities $(1)-(10)$ via known methods (including the WZ method and hypergeometric series identies)?

Your comments are welcome!

Recently, I found the following three (conjectural) identities for $\pi^2$: $$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3,\tag{1}$$ $$\sum_{k=1}^\infty\frac{(42k^2-23k+3)16^k}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^2}2,\tag{2}$$ $$\sum_{k=1}^\infty\frac{(92k^2-61k+9)64^k}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=8\pi^2.\tag{3}$$ The converging rates of the series in $(1)$-$(3)$ are $4/729$, $1/64$ and $4/27$, respectively. One can easily check the identities $(1)$-$(3)$ numerically.

Let $P(k)=145k^2-104k+18$. Motivated by $(1)$, I also conjecture the following identities involving harmonic numbers:

$$\sum_{k=1}^\infty\frac{6P(k)(H_{3k-1}-H_{k-1})-232k+89}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=18\zeta(3),\tag{4}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{2k-1}-H_{k-1})-\frac{3(58k^2-40k+7)}{2(2k-1)}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\zeta(3),\tag{5}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{3k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{17k+32}{9k}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{7\pi^4}{180},\tag{6}$$ $$\sum_{k=1}^\infty\frac{P(k)(297H_{3k-1}^{(2)}-192H_{2k-1}^{(2)}-978H_{k-1}^{(2)})+\frac{27(180k^2+12k-35)}{(2k-1)^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{167}{20}\pi^4,\tag{7}$$ and $$\sum_{k=1}^\infty\frac{P(k)H_{k-1}^{(3)}+\frac{28(2k-1)}{17k^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{528\zeta(5)-46\pi^2\zeta(3)}{17},\tag{8}$$ where $H_n:=\sum_{0<j\le n}\frac1j$ and $H_n^{(m)}:=\sum_{0<j\le n}\frac1{j^m}$ for $m=2,3,\ldots$.

Let $Q(k)=(3k-1)(14k-3)=42k^2-23k+3$. Motivated by $(2)$, I conjecture the following identities: $$\sum_{k=1}^\infty\frac{16^k\left(6Q(k)(H_{2k-1}-H_{k-1})-\frac{196k^2-100k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}62}=2\pi^2\log2-7\zeta(3),\tag{9}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(H_{4k-1}-H_{2k-1})-\frac{28k^2-76k+19}{12(2k-1)}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}62}=\frac{2\pi^2\log2+35\zeta(3)}{12},\tag{10}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-H_{2k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{6k+1}{k}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}62}=\frac{5\pi^4}{24},\tag{11}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-5H_{2k-1}^{(2)}-3H_{k-1}^{(2)})+\frac{32k(3k-1)}{(2k-1)^2}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}62}=\frac{\pi^4}{6}.\tag{12}$$

Let $R(k)=92k^2-61k+9$. Inspired by $(3)$, I also conjecture the following identities: $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{3k-1}-5H_{k-1})-60k+21\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=48\pi^2\log2-56\zeta(3),\tag{13}$$ $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{2k-1}-2H_{k-1})-\frac{124k^2-79k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=112\zeta(3),\tag{14}$$ and $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(6H_{4k-1}-H_{k-1})-\frac{344k^2-281k+59}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=560\zeta(3).\tag{15}$$

QUESTION. Can one prove the new identities $(1)-(15)$ via known methods (including the WZ method and hypergeometric series identies)?

Your comments are welcome!